## Complex Numbers Quiz-7

As per analysis for previous years, it has been observed that students preparing for JEE MAINS find Mathematics out of all the sections to be complex to handle and the majority of them are not able to comprehend the reason behind it. This problem arises especially because these aspirants appearing for the examination are more inclined to have a keen interest in Mathematics due to their ENGINEERING background.

Furthermore, sections such as Mathematics are dominantly based on theories, laws, numerical in comparison to a section of Engineering which is more of fact-based, Physics, and includes substantial explanations. By using the table given below, you easily and directly access to the topics and respective links of MCQs. Moreover, to make learning smooth and efficient, all the questions come with their supportive solutions to make utilization of time even more productive. Students will be covered for all their studies as the topics are available from basics to even the most advanced.

Q1. If the roots of the equation 1/(x+p)+1/(x+q)=1/r are equal in magnitude but opposite in sign, then the product of the roots will be
•  (p2+q2)/2
•  -((p2+q2))/2
•  (p2-q2)/2
•  -(p2-q2)/2
Solution
(b) Given equation 1/(x+p)+1/(x+q)=1/r can be rewritten as x2+x(p+q-2r)+pq-pr-qr=0 …(i)
Let roots are Î± and-Î±, then the product of roots -Î±2=pq-pr-qr-r(p+q) ….(ii)
and sum of roots, 0=-(p+q-2r) ⇒r=(p+q)/2 ...(iii)
On solving Eqs. (ii) and (iii), we get -Î±2=pq-(p+q)/2(p+q) =-1/2{(p+q)2-2pq} ⇒ Î±2=-((p2+q2))/2

Q2.If Î± and Î² are different complex numbers with |Î²|=1, then |(Î²-Î±)/(1-Î± ̅Î²)| is
•  0
•  3/2
•  1/2
•  1
Solution

Q3.  If Ï‰ is acomplex cube root of unity, then the value of sin⁡{(Ï‰10+Ï‰23 )Ï€-Ï€/6} is
•   1/√2
•  √3/2
•  -1/√2
•  1/2
Solution
(d) Since, Ï‰ is a complex cube root of unity
Now, Ï‰10+Ï‰23=(Ï‰3 )3 Ï‰+(Ï‰3 )7 Ï‰2 =Ï‰+Ï‰2=-1
∴ sin⁡{(Ï‰10+Ï‰23 )Ï€-Ï€/6}=sin⁡(-Ï€-Ï€/6)=sin ⁡Ï€/6=1/2

Q4. If x=3+i, then x3-3x2-8x+15 is equal to
•  45
•  -15
•  10
•  6
Solution
(b) Given, x=3+i …(i)
Now, x3-3x2-8x+15 =(3+i)3-3(3+i)2-8(3+i)+15 =(27+i3+27i+9i2 )-3(9+i2+6i)-24-8i+15 =-15

Q5.If cos⁡Î±+2 cos⁡Î²+3 cos⁡Î³=sin⁡Î±+2 sin⁡Î²+3 sin⁡Î³=0 and Î±+Î²+Î³=n Ï€, then sin⁡3 Î±+8 sin⁡3 Î²+27 sin⁡3Î³=
•  0
•  3
•  8
•  -18
Solution
(a) 0

Q6. If Ï‰=(-1+√3 i)/2, then (3+Ï‰+3Ï‰2 )4 is
•  16
•  -16
• 16Ï‰
•  16Ï‰2
Solution
(c) (3+Ï‰+3Ï‰2 )4=[3+(1+Ï‰2 )+Ï‰]4 =[-3Ï‰+Ï‰]4 =(-2Ï‰)4 =16Ï‰

Q7.For any complex number z, the minimum value of |z|+|z-2i|, is
•  0
•  1
•  2
•  4
Solution
(c) Using triangle inequality, we have |z-2i|≥|2i|-|z|⇒|z-2i|+|z|≥2 Hence, the minimum value of |z-2i|+|z| is 2

Q8.If |z1 |=|z2 |=|z3 | and z1+z2+z3=0, then z1,z2,z3 are vertices of
•  A right angled triangle
•  An equilateral triangle
•  Isosceles triangle
•  Scalene triangle
Solution
(b) We have, |z1 |=|z2 |=|z3 | =1
⇒ Origin is the circumcentre of the triangle with the circum radius 1
Also, z1+z2+z3=0 ⇒(z1+z2+z3)/3=0
⇒ Centroid coincides with the origin Hence, the circumcenter and centroid coincides Consequently the triangle is equilateral

Q9.Consider the following statements: 1. If the quadratic equation is ax2+bx+c=0 such that a+b+c=0, then roots of the equation ax2+ bx+c=0 will be 1, c/a. 2. If ax2+bx+c=0 is quadratic equation such that a-b+c=0, then roots of the equation will be, -1,c/a. Which of the statements given above are correct?
•  Only (1)
•  Only (2)
•  Both (1) and (2)
•  Neither (1) nor (2)
Solution
(a) Only (1)

Q10. The number which exceeds its positive square roots by 12, is
•  9
•  16
•  25
• None of these
Solution
(b) Let the required number is x.
According to given condition x=√x+12
⇒x-12=√x
⇒x2-25x+144=0
⇒x2-16x-9x+144=0
⇒x=16,9 Since x=9 does not hold the condition.
∴x=16

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